Question

If tanQ=3/4 find the value of cos2Q-sin2Q

Answer 1

Answer:

here is your answer

Step-by-step explanation:

sin(θ) = 2/5 , 90° ≤ θ ≤ 180°.  

sin(β) = 1/2 , 0° ≤ β ≤ 90°.  

1) sin²(β) + cos²(β) = 1  

cos²(β) = 1 - sin²(β)  

= 1 - (1/2)² = 3/4  

cos(β) = ±½√3 . Note however that cos is positive on [0, 90> , so  

cos(β) = ½√3  

2)  

sec²(β) ≡ 1/cos²(β) = 4/3  

3)  

tan(θ) = sin(θ)/cos(θ)  

= sin(θ) / ( - √(1-sin²(θ) ) [minus sign because cos is negative on interval <90,180>]  

= (2/5) / ( - √( 1 - 4/25) )  

= -2 / √21  

4)  

cosec(β) + tan(90-θ) =  

1/sin(β) + 1/tan(θ) =  

1/ (1/2) + (-(√21) /2) =  

2 - ½√21